English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(3x)-3cos(3x)+1=0,0<= x<= 180

Solution

2 sin^{2} (3 x) - 3 cos (3 x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Solution

x = \frac{0.81462 \dots}{3}, x = \frac{36 0 ^{\circ} - 0.81462 \dots}{3}, x = \frac{0.81462 \dots + 36 0 ^{\circ}}{3}

Radians

x = \frac{0.81462 \dots}{3}, x = \frac{2 π - 0.81462 \dots}{3}, x = \frac{0.81462 \dots + 2 π}{3}

Solution steps

2 sin^{2} (3 x) - 3 cos (3 x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Rewrite using trig identities

1 + 2 sin^{2} (3 x) - 3 cos (3 x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + 2 (1 - cos^{2} (3 x)) - 3 cos (3 x)

Simplify

1 + 2 (1 - cos^{2} (3 x)) - 3 cos (3 x) : - 2 cos^{2} (3 x) - 3 cos (3 x) + 3

1 + 2 (1 - cos^{2} (3 x)) - 3 cos (3 x)

Expand

2 (1 - cos^{2} (3 x)) : 2 - 2 cos^{2} (3 x)

2 (1 - cos^{2} (3 x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (3 x)

= 2 \cdot 1 - 2 cos^{2} (3 x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (3 x)

= 1 + 2 - 2 cos^{2} (3 x) - 3 cos (3 x)

Add the numbers:

1 + 2 = 3

= - 2 cos^{2} (3 x) - 3 cos (3 x) + 3

= - 2 cos^{2} (3 x) - 3 cos (3 x) + 3

3 - 2 cos^{2} (3 x) - 3 cos (3 x) = 0

Solve by substitution

3 - 2 cos^{2} (3 x) - 3 cos (3 x) = 0

Let:

cos (3 x) = u

3 - 2 u^{2} - 3 u = 0

3 - 2 u^{2} - 3 u = 0 : u = - \frac{3 + 33}{4}, u = \frac{33 - 3}{4}

3 - 2 u^{2} - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} - 3 u + 3 = 0

Solve with the quadratic formula

- 2 u^{2} - 3 u + 3 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 3, c = 3

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 3 = 33

(- 3)^{2} - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 2 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 3^{2} + 24

3^{2} = 9

= 9 + 24

Add the numbers:

9 + 24 = 33

= 33

u_{1, 2} = \frac{- ( - 3 ) \pm 33}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 33}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 33}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 33}{2 ( - 2 )} : - \frac{3 + 33}{4}

\frac{- ( - 3 ) + 33}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 + 33}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 + 33}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{3 + 33}{4}

u = \frac{- ( - 3 ) - 33}{2 ( - 2 )} : \frac{33 - 3}{4}

\frac{- ( - 3 ) - 33}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 - 33}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 - 33}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

3 - 33 = - (33 - 3)

= \frac{33 - 3}{4}

The solutions to the quadratic equation are:

u = - \frac{3 + 33}{4}, u = \frac{33 - 3}{4}

Substitute back

u = cos (3 x)

cos (3 x) = - \frac{3 + 33}{4}, cos (3 x) = \frac{33 - 3}{4}

cos (3 x) = - \frac{3 + 33}{4}, cos (3 x) = \frac{33 - 3}{4}

cos (3 x) = - \frac{3 + 33}{4}, 0 \leq x \leq 18 0^{\circ} :

No Solution

cos (3 x) = - \frac{3 + 33}{4}, 0 \leq x \leq 18 0^{\circ}

- 1 \leq cos (x) \leq 1

No Solution

cos (3 x) = \frac{33 - 3}{4}, 0 \leq x \leq 18 0^{\circ} : x = \frac{arccos ( \frac{33 - 3}{4} )}{3}, x = \frac{36 0 ^{\circ} - arccos ( \frac{33 - 3}{4} )}{3}, x = \frac{arccos ( \frac{33 - 3}{4} ) + 36 0 ^{\circ}}{3}

cos (3 x) = \frac{33 - 3}{4}, 0 \leq x \leq 18 0^{\circ}

Apply trig inverse properties

cos (3 x) = \frac{33 - 3}{4}

General solutions for

cos (3 x) = \frac{33 - 3}{4}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

3 x = arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n, 3 x = 36 0^{\circ} - arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n

3 x = arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n, 3 x = 36 0^{\circ} - arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n

Solve

3 x = arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n : x = \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

3 x = arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n

Divide both sides by

3

3 x = arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n

Divide both sides by

3

\frac{3 x}{3} = \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

Simplify

x = \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

x = \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

Solve

3 x = 36 0^{\circ} - arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n : x = 12 0^{\circ} - \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

3 x = 36 0^{\circ} - arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n

Divide both sides by

3

3 x = 36 0^{\circ} - arccos (\frac{33 - 3}{4}) + 36 0^{\circ} n

Divide both sides by

3

\frac{3 x}{3} = 12 0^{\circ} - \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

Simplify

x = 12 0^{\circ} - \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

x = 12 0^{\circ} - \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

x = \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}, x = 12 0^{\circ} - \frac{arccos ( \frac{33 - 3}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = \frac{arccos ( \frac{33 - 3}{4} )}{3}, x = \frac{36 0 ^{\circ} - arccos ( \frac{33 - 3}{4} )}{3}, x = \frac{arccos ( \frac{33 - 3}{4} ) + 36 0 ^{\circ}}{3}

Combine all the solutions

x = \frac{arccos ( \frac{33 - 3}{4} )}{3}, x = \frac{36 0 ^{\circ} - arccos ( \frac{33 - 3}{4} )}{3}, x = \frac{arccos ( \frac{33 - 3}{4} ) + 36 0 ^{\circ}}{3}

Show solutions in decimal form

x = \frac{0.81462 \dots}{3}, x = \frac{36 0 ^{\circ} - 0.81462 \dots}{3}, x = \frac{0.81462 \dots + 36 0 ^{\circ}}{3}

Popular Examples

2sin(x/2)=1-cos(x)

2 sin (\frac{x}{2}) = 1 - cos (x)

sin(x)=(-sin(2x))/2

sin (x) = \frac{- sin ( 2 x )}{2}

12sin(A)+4=4sin(A)+4

12 sin (A) + 4 = 4 sin (A) + 4

2sin^2(x)=3-3cos(x)

2 sin^{2} (x) = 3 - 3 cos (x)

cos(x)+sin(x)tan(x)=1

cos (x) + sin (x) tan (x) = 1